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Ellipse question

2016 · Shift 2 · Q29
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  5. /2016 · Shift 2 · Q29

Ellipse question

2016 · Shift 2 · Q29

JEE AdvancedMathematicsEllipseMCQ+3 / −1
Let F1(x1,0){F_1}\left( {{x_1},0} \right)F1​(x1​,0) and F2(x2,0){F_2}\left( {{x_2},0} \right)F2​(x2​,0) for x1<0{{x_1} \lt 0}x1​<0 and x2>0{{x_2} \gt 0}x2​>0, be the foci of the ellipse x29+y28=1{{{x^2}} \over 9} + {{{y^2}} \over 8} = 19x2​+8y2​=1. Suppose a parabola having vertex at the origin and focus at F2{F_2}F2​ intersects the ellipse at point MMM in the first quadrant and at point NNN in the fourth quadrant. The orthocentre of the triangle F1MN{F_1}MNF1​MN is
  1. A
    (−910,0)\left( { - {9 \over {10}},0} \right)(−109​,0)
  2. B
    (23,0)\left( { {2 \over {3}},0} \right)(32​,0)
  3. C
    (910,0)\left( { {9 \over {10}},0} \right)(109​,0)
  4. D
    (23,6)\left( {{2 \over 3},\sqrt 6 } \right)(32​,6​)
View written solutionFree

Correct answer: A

  1. Find the foci of the ellipse

The ellipse is

x29+y28=1\frac{x^2}{9}+\frac{y^2}{8}=19x2​+8y2​=1

which is in the standard form

x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1

with

a2=9,b2=8.a^2=9,\quad b^2=8.a2=9,b2=8.

So

a=3,b=22.a=3,\quad b=2\sqrt2.a=3,b=22​.

For an ellipse, the focal distance is

c=a2−b2=9−8=1.c=\sqrt{a^2-b^2}=\sqrt{9-8}=1.c=a2−b2​=9−8​=1.

Hence the foci are

F1=(−1,0),F2=(1,0).F_1=(-1,0),\quad F_2=(1,0).F1​=(−1,0),F2​=(1,0).

So,

x1=−1,x2=1.x_1=-1,\quad x_2=1.x1​=−1,x2​=1.
  1. Equation of the parabola

The parabola has vertex at the origin and focus at F2=(1,0)F_2=(1,0)F2​=(1,0).

A parabola with vertex at (0,0)(0,0)(0,0) and focus (a,0)(a,0)(a,0) has equation

y2=4ax.y^2=4ax.y2=4ax.

Here a=1a=1a=1, so the parabola is

y2=4x.y^2=4x.y2=4x.
  1. Find the intersection points with the ellipse

We solve simultaneously:

x29+y28=1\frac{x^2}{9}+\frac{y^2}{8}=19x2​+8y2​=1

and

y2=4x.y^2=4x.y2=4x.

Substitute y2=4xy^2=4xy2=4x into the ellipse:

x29+4x8=1.\frac{x^2}{9}+\frac{4x}{8}=1.9x2​+84x​=1.

So,

x29+x2=1.\frac{x^2}{9}+\frac{x}{2}=1.9x2​+2x​=1.

Multiply by 181818:

2x2+9x−18=0.2x^2+9x-18=0.2x2+9x−18=0.

Factor:

2x2+12x−3x−18=02x^2+12x-3x-18=02x2+12x−3x−18=0 2x(x+6)−3(x+6)=02x(x+6)-3(x+6)=02x(x+6)−3(x+6)=0 (x+6)(2x−3)=0.(x+6)(2x-3)=0.(x+6)(2x−3)=0.

Thus,

x=−6orx=32.x=-6 \quad \text{or} \quad x=\frac32.x=−6orx=23​.

Since the ellipse has ∣x∣≤3|x|\le 3∣x∣≤3, x=−6x=-6x=−6 is invalid. Hence

x=32.x=\frac32.x=23​.

Then

y2=4x=4⋅32=6y^2=4x=4\cdot \frac32=6y2=4x=4⋅23​=6

so

y=±6.y=\pm \sqrt6.y=±6​.

Therefore the intersection points are

M(32,6),N(32,−6).M\left(\frac32,\sqrt6\right),\qquad N\left(\frac32,-\sqrt6\right).M(23​,6​),N(23​,−6​).
  1. Coordinates of triangle F1MNF_1MNF1​MN

The triangle has vertices

F1=(−1,0),M(32,6),N(32,−6).F_1=(-1,0),\quad M\left(\frac32,\sqrt6\right),\quad N\left(\frac32,-\sqrt6\right).F1​=(−1,0),M(23​,6​),N(23​,−6​).

Notice that MNMNMN is a vertical line:

x=32.x=\frac32.x=23​.

Hence the altitude from F1F_1F1​ to MNMNMN is the horizontal line

y=0.y=0.y=0.

So the orthocentre must lie on the xxx-axis.

  1. Find another altitude

First compute the slope of side F1NF_1NF1​N:

mF1N=−6−032−(−1)=−652=−265.m_{F_1N}=\frac{-\sqrt6-0}{\frac32-(-1)}=\frac{-\sqrt6}{\frac52}=-\frac{2\sqrt6}{5}.mF1​N​=23​−(−1)−6​−0​=25​−6​​=−526​​.

So the altitude from MMM has slope equal to the negative reciprocal:

m⊥=526.m_{\perp}=\frac{5}{2\sqrt6}.m⊥​=26​5​.

Equation of altitude through M(32,6)M\left(\frac32,\sqrt6\right)M(23​,6​) is

y−6=526(x−32).y-\sqrt6=\frac{5}{2\sqrt6}\left(x-\frac32\right).y−6​=26​5​(x−23​).

Since orthocentre lies on y=0y=0y=0, put y=0y=0y=0:

−6=526(x−32).-\sqrt6=\frac{5}{2\sqrt6}\left(x-\frac32\right).−6​=26​5​(x−23​).

Multiply both sides by 262\sqrt626​:

−12=5(x−32).-12=5\left(x-\frac32\right).−12=5(x−23​).

Thus,

x−32=−125x-\frac32=-\frac{12}{5}x−23​=−512​ x=32−125=15−2410=−910.x=\frac32-\frac{12}{5}=\frac{15-24}{10}=-\frac{9}{10}.x=23​−512​=1015−24​=−109​.

So the orthocentre is

(−910,0).\left(-\frac{9}{10},0\right).(−109​,0).
  1. Check with options

This matches:

(−910,0)\boxed{\left(-\frac{9}{10},0\right)}(−109​,0)​

which is Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

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